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3 years ago

2NBr3 + 3NaOH – N2 + 3NaBr + 3HBrO

Chemistry

1 answer:

Damm [24]3 years ago

3 0

Answer:

50 formula units of NBr3 will yield 75 formula units of NaBr

57 formula units of NaOH will yield 57 formula units of NaBr

Explanation:

The equation of the reaction is ;

2NBr3 + 3NaOH – N2 + 3NaBr + 3HBrO

Number of formula units of NaBr formed when 50NBr3 reacts was obtained from;

2 formula units of NBr3 yields 3 formula units of NaBr

50 formula units of NBr3 will yield 50 ×3/2 = 75 formula units of NaBr

Hence 50 formula units of NBr3 will yield 75 formula units of NaBr

Then;

Since 3 formula units of NaOH yields 3 formula units of NaBr

57 formula units of NaOH will yied 57×3/3 = 57 formula units of NaBr

Hence 57 formula units of NaOH will yield 57 formula units of NaBr

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galben [10]

Answer:

Danger

Explanation:

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2 years ago

Convert 15.2 moles of K to atoms of K.

choli [55]

Answer:

$\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}$

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

$\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

Multiply by the given number of moles: 15.2

$15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

The moles of potassium cancel.

$15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }$

The denominator of 1 can be ignored.

$15.2 * {6.022*10^{23} \ atoms \ K }{$

Multiply.

$9.15344*10^{24} \ atoms \ K$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

$9.15*10^{24} \ atoms \ K$

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

3 0

3 years ago

A 0.288 g sample of an unknown monoprotic acid is dissolved in water and titrated with a 0.115 M NaOH solution. After the additi

Sholpan [36]

Answer:

74.0 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

At the equivalence point, 33.83 mL of 0.115 M NaOH react.

0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol

Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.

Step 4: Calculate the molar mass of the acid

3.89 × 10⁻³ moles of HA have a mass of 0.288 g.

M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol

4 0

3 years ago

Please help Asap………………

vladimir2022 [97]

Answer:

3rd

Explanation:

6 0

3 years ago

Read 2 more answers

How many moles of neon occupy a volume of 14.3 l at stp? how many moles of neon occupy a volume of 14.3 l at stp? 1.57 moles 0.6

andriy [413]

The number of neon moles that occupy a volume of 14.3 l at STP is calculated as follows

At STP 1 mole = 22.4 liters

what about 14.3 liters

by cross multiplication
= (1 mole x 14.3 l)/22.4 l =0.638 moles of neon

5 0

3 years ago